Can you make the triviality of $\langle a,b,c \mid aba^{-1} = b^2, bcb^{-1} = c^2, cac^{-1} = a^2 \rangle$ more trivial?

I recently learned the following pleasant fact. (It was in the proof of Proposition 3.1 of this paper - but don't worry, there's no model theory in this question.)

Let $$G$$ be a group, and let $$a,b,c\in G$$. If $$aba^{-1} = b^2$$, $$bcb^{-1} = c^2$$, and $$cac^{-1} = a^2$$, then $$a = b = c = e$$. Put another way, the group defined by generators and relations $$\langle a,b,c \mid aba^{-1} = b^2, bcb^{-1} = c^2, cac^{-1} = a^2 \rangle$$ is the trivial group.

I came up with the following elementary, but ugly, proof:

The relations can be rewritten as (1) $$ab = b^2a$$, (2) $$bc = c^2b$$, (3) $$ca = a^2c$$.

Using (1), (2), and (3), we can rewrite $$a^4bc = a^4c^2b = c^2ab = c^2b^2a$$.

But we can also rewrite $$a^4bc = b^{16}a^4c = b^{16}ca^2 = c^{2^{16}}b^{16}a^2$$.

So $$c^{2^{16}}b^{16}a^2 = c^2b^2a$$. This implies $$a = b^{-16}c^{2-2^{16}}b^2$$.

Substituting for $$a$$ in (1) above, $$b^{-16}c^{2(1-2^{15})}b^3 = b^{-14}c^{2(1-2^{15})}b^2$$, and cancelling from both sides, $$c^{2(1-2^{15})}b = b^{2}c^{2(1-2^{15})}$$.

But now by (2), we have $$bc^{1-2^{15}} = b^{2}c^{2(1-2^{15})}$$, and $$b = c^{2^{15}-1}$$. But then $$b$$ and $$c$$ commute, so $$bcb^{-1} = c^2$$ implies $$c = c^2$$, and $$c = e$$. It then follows easily that $$a = b = c = e$$.

Question: Is there a better way to see this? i.e. a more abstract proof, or at least one that doesn't involve manipulating words of length $$2^{16}$$?

• A slightly relevant remark, if there are four generators with similar relations, the group will be infinite. That is, $⟨a,b,c,d∣aba^{−1}=b^2, bcb^{−1}=c^2,cdc^{−1}=d^2, dad^{−1}=a^2⟩$ is not finite. c.f. Serre's Trees Page 9. – userabc Jan 4 at 5:02
• @userabc Thanks for the comment - this is actually the very next remark in the paper I linked to. But the authors just call it a "well-known fact", so it's nice to have the reference to Serre's book. And I see that my question is Exercise 1) on p. 10 of Trees. – Alex Kruckman Jan 4 at 5:14
• ... and now that I can search for an exercise number in a well-known book, I find that there are a number of questions on this site about the same group. For example, Jim Belk gave almost an identical proof to the one I found here, and Martin Brandenburg asked essentially the same question I'm asking here. Together, this evidence makes me think that there's unlikely to be a nicer proof. It would be reasonable to close this question as a duplicate. – Alex Kruckman Jan 4 at 5:26
• There is another proof here. – Derek Holt Jan 4 at 8:58
• Once you have proved that $a \in \langle b,c \rangle$, you can use the more conceptual argument given by Bhaskar Vashishth in the linked proof, that $G$ is perfect, but $G = \langle b,c \rangle$ is solvable, so $G$ is trivial. – Derek Holt Jan 4 at 15:05